How do you graph #f(x)=(1-x^2)/x# using holes, vertical and horizontal asymptotes, x and y intercepts?

1 Answer
Jul 6, 2018

no holes
vertical asymptote at #x = 0#
#x#-intercepts at #(-1, 0), (1, 0)#
#y#-intercept is undefined
no horizontal asymptote
slant/oblique asymptote at #y = -x#

Explanation:

Given: #f(x) = (1-x^2)/x#

This is a rational function: #(N(x))/(D(x)) = (a_nx^n+...)/(b_mx^m+...)#

Factor the numerator:
#f(x) = -(x^2 -1)/x = -((x-1)(x+1))/x#

Holes: Since there are no terms that cancel, there are no holes.

Find vertical asymptotes: when #D(x) = 0 => x = 0#

Find x-intercepts by setting #f(x) = 0#:

#0 = (1-x^2)/x#

#0 * x = (1 - x^2)#

#1 - x^2 = 0#

#x^2 = 1#

#x = +- 1#

#x#-intercepts at #(-1, 0), (1, 0)#

Find y-intercept by setting #x = 0 => y = #undefined

Find horizontal asymptote by comparing #m & n#:

If #n < m#, horizontal asymptote is #y = 0#

If #n = m#, horizontal asymptote is #y = a_n/b_m#

If #n > m# there is no horizontal aymptote

In the given equation, #n = 2, m = 1 => n > m# which means there is no horizontal asymptote.

If #n = m+1# there is a slant /oblique asymptote.

You need to use long division to find the slant/oblique asymptote :

#" "-x#
#x |bar(-x^2 + 0x + 1)#
#" "ul(-x^2)#
#" "0x + 1#

#" "color(red)(-x) + 1/x#
#x |bar(-x^2 + 0x + 1)#
#" "ul(-x^2)#
#" "0x + 1# This is the remainder

slant/oblique asymptote is at #color(red)(y = -x)#